Question

Find the sum of the first (i) $75$ positive integers (ii) $125$ natural numbers

Answer 1

(i) The arithmetic series of first $75$ positive integers is $1+2+3+.....+75$ where $a_1=1,a_2=2$ and $n=75$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=2-1=1$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now substitute $n=75,a=1$ and $d=1$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{75}=\frac{75}{2}\left[\right(2\times 1)+(75-1\left)1\right]=\frac{75}{2}(2+74)=\frac{75}{2}\times 76=75\times 38=2850$

Hence, the sum of first $75$ positive integers is $2850$.

(ii) The arithmetic series of first $125$ natural numbers is $1+2+3+.....+125$ where $a_1=1,a_2=2$ and $n=125$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=2-1=1$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now substitute $n=125,a=1$ and $d=1$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{125}=\frac{125}{2}\left[\right(2\times 1)+(125-1\left)1\right]=\frac{125}{2}(2+124)=\frac{125}{2}\times 126=125\times 63=7875$

Hence, the sum of first $125$ natural numbers is $7875$.